Question

高一数学直线与圆方程在直线x-y+2根号2=0上求一点p,使p到圆x^2+y^2=1的切线长最短,和此时切线的长.过程详

Answer 1

设P=(XP,YP)

P到圆的切线长为

√(XP^2+YP^2-1)

P在x-y+2(√2)=0上

YP=XP+2(√2)

√(XP^2+YP^2-1)

=√{XP^2+[XP+2(√2)]^2-1}

=√[2XP^2+2(√2)XP+7]

=√{2[XP+(√2)/2]^2+6}

[XP+(√2)/2]^2>=0

当[XP+(√2)/2]^2=0时切线最短

切线长度为√6